Page:Annales de mathématiques pures et appliquées, 1823-1824, Tome 14.djvu/178

$\left(1+q^{2}\right)\left({\frac {P'-p}{Q'-q}}\right)^{2}-2pq\left({\frac {P'-p}{Q'-q}}\right)+\left(1+p^{2}\right)=$

$\left(1+q^{2}\right)\left({\frac {P''-p}{Q''-q}}\right)^{2}-2pq\left({\frac {P''-p}{Q''-q}}\right)+\left(1+p^{2}\right)$

ou bien

${\frac {\left(1+q^{2}\right)\left({\frac {P'-p}{Q'-q}}\right)^{2}-2pq\left({\frac {P'-p}{Q'-q}}\right)+\left(1+p^{2}\right)}{\left(1+q^{2}\right)\left({\frac {P''-p}{Q''-q}}\right)^{2}-2pq\left({\frac {P''-p}{Q''-q}}\right)+\left(1+p^{2}\right)}}=1.$

d’où

$\left({\frac {Q'-q}{Q''-q}}\right)^{2}.{\frac {\left(1+q^{2}\right)\left({\frac {P'-p}{Q'-q}}\right)^{2}-2pq\left({\frac {P'-p}{Q'-q}}\right)+\left(1+p^{2}\right)}{\left(1+q^{2}\right)\left({\frac {P''-p}{Q''-q}}\right)^{2}-2pq\left({\frac {P''-p}{Q''-q}}\right)+\left(1+p^{2}\right)}}$

$=\left({\frac {Q'-q}{Q''-q}}\right)^{2}=\left({\frac {P'-p}{P''-p}}\right)^{2}$

ou, en exécutant la multiplication dans le premier membre,

${\frac {\left(1+q^{2}\right)(P'-p)^{2}-2pq(P'-p)(Q'-q)+\left(1+p^{2}\right)(Q'-q)^{2}}{\left(1+q^{2}\right)(P''-p)^{2}-2pq(P''-p)(Q''-q)+\left(1+p^{2}\right)(Q''-q)^{2}}}$

$=\left({\frac {Q'-q}{Q''-q}}\right)^{2}=\left({\frac {P'-p}{P''-p}}\right)^{2}$

ou encore

${\frac {\left(1+p^{2}+q^{2}\right)\left(1+P'^{2}+Q'^{2}\right)-(1+pP'+qQ')^{2}}{\left(1+p^{2}+q^{2}\right)\left(1+P''^{2}+Q''^{2}\right)-(1+pP''+qQ'')^{2}}}$

$=\left({\frac {Q'-q}{Q''-q}}\right)^{2}=\left({\frac {P'-p}{P''-p}}\right)^{2}\,;$

mais l’équation (II) peut être écrite ainsi

${\frac {\left(1+p^{2}+q^{2}\right)\left(1+P'^{2}+Q'^{2}\right)-(1+pP'+qQ')^{2}}{\left(1+p^{2}+q^{2}\right)\left(1+P''^{2}+Q''^{2}\right)-(1+pP''+qQ'')^{2}}}={\frac {\lambda '^{2}\left(1+P'^{2}+Q'^{2}\right)}{\lambda ''^{2}\left(1+P''^{2}+Q''^{2}\right)}}\,;$

donc

$\left({\frac {Q'-q}{Q''-q}}\right)^{2}=\left({\frac {P'-p}{P''-p}}\right)^{2}={\frac {\lambda '^{2}\left(1+P'^{2}+Q'^{2}\right)}{\lambda ''^{2}\left(1+P''^{2}+Q''^{2}\right)}}\,;$

c’est-à-dire,